Work done by a conservative force

A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Equivalently, if a particle travels in a closed loop, the net work done (the sum of the force acting along the path multiplied by the distance travelled) by a conservative force is zero.

Does the work done by the force remain 0 even if it varies at all points on the loop? From the definition given in wikipedia it seems as if it is defined for work done by a constant conservative force. Is the work done over a loop 0 for variable forces as well? Can it be somehow proven for a variable force?

Does the work done by the force remain 0 even if it varies at all points on the loop ?

Yes. For example, the gravitational force.

Note, that in general fields are not conservative. So if you write an arbitrary force, the work will not be zero.

• No I mean varying with point and for gravitational force we can take approximation that it is constant. I know the work is 0 even for forces varying extremely from point to point over a loop too, but can we just apply it everywhere ? How ? – Rijul Gupta Nov 23 '13 at 18:31
• physics.stackexchange.com/questions/87621/… please take a look and tell me what I am doing wrong here then ? – Rijul Gupta Nov 23 '13 at 18:48
• The field is electrostatic, I think it should be conservative only – Rijul Gupta Nov 23 '13 at 19:11
• So if I set up an electric field which has a dependency on y as I mentioned then in that case it is possible that the field may not be conservative even though it is electrostatic in nature ? – Rijul Gupta Nov 23 '13 at 19:32
• But just in case hypothetically, I did what will be its consequence ? – Rijul Gupta Nov 23 '13 at 19:40

The work done by a force along a path $\gamma$ is defined as $$W = \int_\gamma \vec{F}(\vec{r})\cdot d\vec{r} = \int_a^b \vec{F}(\gamma(t))\cdot\dot{\gamma}(t)\,dt$$ where the last equality is the actual definition of the line-integral of a vectorfield along a curve. Note that the force is explicitly depending on position. As you state correctly, a force is called conservative iff the work done is a function of start- and end-point of the path only: $$F\, cons. \Leftrightarrow W = V[\gamma(a)]-V[\gamma(b)] \Leftrightarrow F=-\nabla V \Rightarrow \nabla\times F=0$$

The last implication is an equivalence if the domain on which $F$ is defined, is star-shaped. It is a special case of the Poincaré Lemma.

Maxwells equations tell us, that in the absence of time-variing magnetic fields, the electric field is curl-free and hence posesses a potential $$\nabla\times E=0 \Rightarrow E=-\nabla V$$ Hence every electrostatic field is conservative, which can easily be checked by calculating the curl!

Sidenote (Please correct me if i that is not on spot):
You may argue, that the electrostatic field of a point-charge $E=-q\,\hat{r}/r^2$ is singular at the origin and the curl-criterion therfore not strictly applicable. However this is kind of an artifact from modeling the chargedistribution as a Dirac Delta $\rho=q\,\delta(\vec{r})$ infinitly narrow peaked in space. One may instead think of 'smearing out' the singular distribution a bit in space, so everything becomes nice and smooth, making Poincarés Lemma applicable.

protected by Qmechanic♦Dec 28 '16 at 6:05

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